在PHP中通过引用传递数据库连接 [英] Passing database connection by reference in PHP

问题描述

问题是如果一个数据库连接应该通过引用或按值传递？

The question is if a database connection should be passed in by reference or by value?

对我来说，我特别问一个PHP到MySQL的连接，认为它适用于所有数据库。

For me I'm specifically questioning a PHP to MySQL connection, but I think it applies to all databases.

我听说在PHP中传递一个变量到一个函数或对象，它被复制在内存中，因此使用两次多记忆。我还听说，只有在对值进行更改（例如，从数组中添加/删除一个键）后才会复制。

I have heard that in PHP when you pass a variable to a function or object, that it is copied in memory and therefore uses twice as much memory immediately. I have also heard that it's only copied once changes have been made to the value (such as a key being added/removed from an array).

在数据库连接中，我认为它是在函数内改变，因为查询可以改变像最后一个插入id或num行。 （我想这是另一个问题：像num行和插入id存储在连接或实际调用回到数据库的事情）。

In a database connection, I would think it's being changed within the function as the query could change things like the last insert id or num rows. (I guess this is another question: are things like num rows and insert id stored within the connection or an actual call is made back to the database?)

如果连接通过引用或值传递，它是内存或速度明智？是否有区别PHP 4 vs 5？

So, does it matter memory or speed wise if the connection is passed by reference or value? Does it make a difference PHP 4 vs 5?

// $connection is resource
function DoSomething1(&$connection) { ... }
function DoSomething2($connection) { ... }

推荐答案

PHP资源是一种特殊类型，它本身就是一个引用。通过值或通过引用显式传递它不会产生差异（即，它仍然是引用）。你可以在PHP4下自己检查：

A PHP resource is a special type that already is a reference in itself. Passing it by value or explicitly by reference won't make a difference (ie, it's still a reference). You can check this for yourself under PHP4:

function get_connection() {
  $test = mysql_connect('localhost', 'user', 'password');
  mysql_select_db('db');
  return $test;
}

$conn1 = get_connection();
$conn2 = get_connection(); // "copied" resource under PHP4

$query = "INSERT INTO test_table (id, field) VALUES ('', 'test')";
mysql_query($query, $conn1);
print mysql_insert_id($conn1)."<br />"; // prints 1

mysql_query($query, $conn2);
print mysql_insert_id($conn2)."<br />"; // prints 2

print mysql_insert_id($conn1); // prints 2, would print 1 if this was not a reference

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